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December 12th, 2017, 07:19 AM  #1 
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 15 Thanks: 0 Math Focus: Calculus  How to solve this problem involving friction?
The problem is as follows: In the figure shown, Calculate the Force labeled as $F$ which must be applied to the wooden block $B$ whose mass is $10\textrm{ kg}$ thus it would not slide over the wall. The friction coefficient in the surface is $\frac{1}{3}$ and gravity acceleration is $g=10\,\frac{m}{s^{2}}$ I thought to solve the problem by summing all forces as explained below: $mg+F\cos37^{o}\mu\times n=0$ The normal in this case is the same as $F\sin37^{o}$ By plugin the values in the equation I got to this: $(10\,kg)(10\frac{m}{s^{2}})+F\cos37^{o}(\frac{1}{3})\times F\sin37^{o}=0$ $(10\,kg)(10\frac{m}{s^{2}})+F(\frac{3}{5})(\frac{1}{3})\times F(\frac{3}{5})=0$ $F(\frac{1}{5})(31)=(10\,kg)(10\frac{m}{s^{2}}$ $F(\frac{2}{5})=100\,kg\times\frac{m}{s^{2}}$ $F=250\,kg\times\frac{m}{s^{2}}$ However I'm confused as why? the Force is negative. 
December 12th, 2017, 08:26 AM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 239 Thanks: 126 Math Focus: Dynamical systems, analytic function theory, numerics 
I can't follow your computations at all but you seem to have a few misconceptions which are noticeable. Likely, these are the reason your computation is not intelligible. 1. There is no reason force can't be negative. 2. Force is a vector. 3. The expression $\mu \times n$ is a vector cross product, not multiplication. 4. You can't do normal algebra on vectors. Specifically, you have $F = ma$ but in this expression $F,a$ are both vectors. So you can't "solve" for $m$ as $m = \frac{F}{a}$. The expression on the right is meaningless. I would suggest you forget this particular problem for the moment and revisit the fundamentals of matrix/vector algebra. 
December 12th, 2017, 08:58 AM  #3  
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 15 Thanks: 0 Math Focus: Calculus  Quote:
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Although I'm aware that this problem can be solved by going on different routes I would prefer to follow the one which avoid the use of calculus and rely on simple algebra.  
December 12th, 2017, 09:55 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,691 Thanks: 1350 
two cases ... 1. If $F\cos(37) < mg$ then the force of static friction acts upward to maintain equilibrium ... $F\cos(37) + f = mg$ note that $f \le \mu \cdot N$, where $N = F\sin(37)$, so the maximum upward force of friction is $\mu F\sin(37)$ which in turn gives a minimum value for $F$ ... $F_{min}\cos(37) + \mu F_{min}\sin(37) = mg \implies F_{min} = \dfrac{mg}{\cos(37) + \mu\sin(37)}$ 2. If $F\cos(37) > mg$ then the force of static friction acts downward to maintain equilibrium ... $F\cos(37)  f = mg$ in this case, if $f = \mu F\sin(37)$, then $F$ is a maximum ... $F_{max}\cos(37)  \mu F_{max}\sin(37) = mg \implies F_{max} = \dfrac{mg}{\cos(37)  \mu\sin(37)}$ therefore, the range of possible values for $F$ for the system to maintain equilibrium is ... $\dfrac{mg}{\cos(37) + \mu\sin(37)} \le F \le \dfrac{mg}{\cos(37)  \mu\sin(37)}$ 
December 12th, 2017, 10:53 AM  #5 
Senior Member Joined: Jun 2015 From: England Posts: 712 Thanks: 203 
This is where you have to use some Physics common sense, even though it conflicts with the maths. No F can't be negative. Not because you can't apply a negative F (you could) or because the maths is wrong. But simply because that would pull the block away from the wall. What the negative sign is telling you is that you have one of the other forces pointing in the wrong direction. Now the weight (force due to gravity on the mass of the block) is fixed in magnitude and direction and you have that correct. You have n, the normal reaction from the wall onto the block going in the wrong direction however, in your free body diagram. Your n is the force exerted by the block on the wall. But your free body diagram is about forces on the block. Once F is fixed, n is fixed (not variable). Now for the difficult one  the friction. The force of friction is variable and only provides enough force to maintain the balance of equilibrium. This could be anything from zero up to the maximum value of limiting friction. Secondly Friction always acts in the opposite direction from the motion (or proposed motion). Now there are two situations. The force F holds the block against the wall with its horizontal component. Thus this equals the reaction between them. But F may either tend to push the block up the wall (in which case friction will act downwards) Or it will not be strong enough to hold the block up and the blcok will tend to slip down the wall. In this case friction will act upwards and combine with (add to) the vertical component of F to support the block in place. I will not repeat the admirable maths that skeeter has already offered, This ramble was by way of explanation. 
December 13th, 2017, 02:02 AM  #6  
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 15 Thanks: 0 Math Focus: Calculus  Quote:
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I also notice that there was an error in vector decomposition. I'm comparing both version in the sketch attached. Quote:
I'm confused about the force F which you say is pushing the block which causes the friction to act downwards. Is this force a maximum or a minimum. Why? The same applies to the force which causes the friction to act upwards. Is the latter a minimum or a maximum? If you could make a drawing explaining this could be perfect. Sorry if my questions do seem silly but I'm in the process of clearing my ideas.  

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